Let $P$ be a point on the plane. Three nonoverlapping equilateral triangles $PA_1A_2$, $PA_3A_4$, $PA_5A_6$ are constructed in a clockwise manner. The midpoints of $A_2A_3$, $A_4A_5$, $A_6A_1$ are $L$, $M$, $N$, respectively. Prove that triangle $LMN$ is equilateral.
Problem
Source: Taiwan 1st TST 2006, 1st independent study, problem 2
Tags: analytic geometry, trigonometry, geometry, geometry proposed
Megus
29.03.2006 15:06
Take an origin of a complex plane at $P$. By smaller letters I'll denote corresponding complex coordinate of a point. Let ${{\epsilon=\cos (\frac{2\pi}{3}})+i \sin (\frac{2\pi}{3}})$. Then: $a_2+a_1 \cdot \epsilon=0$ $a_6+a_5 \cdot \epsilon=0$ $a_4+a_3 \cdot \epsilon=0$. $LMN$ being equilateral is equivalent to: $\frac{1}{2}((a_2+a_3)+(a_1+a_6)\cdot \epsilon+(a_4+a_5)\cdot \epsilon ^2=0$ which is true and easy to check with help of above relations.
darij grinberg
29.03.2006 17:47
k2c901_1 wrote: Let $P$ be a point on the plane. Three nonoverlapping equilateral triangles $PA_1A_2$, $PA_3A_4$, $PA_5A_6$ are constructed in a clockwise manner. The midpoints of $A_2A_3$, $A_4A_5$, $A_6A_1$ are $L$, $M$, $N$, respectively. Prove that triangle $LMN$ is equilateral. See also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16908 post #3. Darij
Virgil Nicula
29.03.2006 19:57
A equivalent enunciation. Let $P$ be a point on the plane of the triangle $ABC$. Three nonoverlapping equilateral triangles $PAA'$, $PBB'$, $PCC'$ are constructed in a clockwise manner. Given are the middlepoints $L$, $M$, $N$ of the segments $A'B$, $B'C$, $C'A$ respectively. Prove that the triangle $LMN$ is equilateral.
Proof. I shall use also the complex numbers. Denote $X(x)\ -$ the point $X$ with the affix $x$ and I shall choose the origin in the point $P$. Thus, $P(0)$, $A(a),\ B(b),\ C(c)$ a.s.o. Define the complex number $\omega =\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}$. Therefore, $\omega ^3=-1$, $\omega ^2=\omega -1$, $\overline {\omega}=\frac{1}{\omega}=-\omega ^2$ and $a'=a\omega$, $b'=b\omega$, $c'=c\omega$ $\Longrightarrow$ $2l=b+a\omega$, $2m=c+b\omega$, $2n=a+c\omega$ $\Longrightarrow$ $2(l-m)=(b-c)+(a-b)\omega$, $2(n-m)=(a-c)+(c-b)\omega$ $\Longrightarrow$ $2(n-m)\omega=$ $(a-c)\omega +(c-b)\omega ^2=$ $(a-c)\omega +(c-b)(\omega -1)=$ $(b-c)+(a-b)\omega =$ $2(l-m)$ $\Longrightarrow$ $\boxed {\ (l-m)=(n-m)\omega\ }$ $\Longleftrightarrow$ $LM=NM$ and $m(\widehat {LMN})=60^{\circ}$, i.e. the triangle $ABC$ is equilateral.
silouan
29.03.2006 21:33
See also here http://www.mathlinks.ro/Forum/viewtopic.php?t=45874 Here in the last post you can find a solution posted by me using pure geometry!!!! What do you think about this ?